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I'm working on this:

Given that [tex]\lim_{n \to \infty} \psi(n)=0[/tex] and that b and c do not depend upon n, prove that:

[tex]\lim_{n\to \infty}\left[ 1+\frac{b}{n} +\frac{\psi(n)}{n}\right]^{cn} = \lim_{n\to\infty} \left(1+\frac{b}{n}\right)^{cn}=e^{bc}[/tex]

So far, I've taken the natural log of both sides, moved the cn into the bottom and applied L'hopitals rule. I get:

[tex]\lim_{n\to\infty}\frac{\frac{1}{1+\frac{b}{n} +\frac{\psi(n)}{n}}\left(\frac{-b}{n^2} + \frac{\psi'(n)}{n}-\frac{\psi(n)}{n^2}\right)}} {\frac{-1}{c n^2}}}[/tex][tex]=\lim_{n\to \infty}bc[/tex]

which breaks down to:

[tex]\lim_{n\to\infty}\frac{1}{1+\frac{b}{n}+\frac{\psi(n)}{n}}\left(-cn\psi'(n)-c\psi(n)+bc)\right)[/tex]

If the limit of a function goes to zero, how do we prove that it's derivative goes to zero?

I'm not sure where to go now, because I don't know what to do with [tex]\psi'(n)[/tex] how can I prove that it's zero? If it IS zero, then the whole thing falls out nicely.

Thanks,

CC

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# Homework Help: Exponential function

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